Is it possible to realize a general compact Riemann surface in $\mathbb CP^2$?

Question

Let $X$ be a compact Riemann surface with smooth boundary $\partial X$. Is it always possible to realize $X$ as a complex submanifold of $\mathbb CP^2$? In other words, is it true that there exists a Riemann surface $Y \subset \mathbb CP^2$ with boundary $\partial Y$ and a bijection $f \colon X \to Y$, $f(\partial X) = \partial Y$ such that $f$ and $f^{-1}$ are both analytic?

Answer 1

a) Every compact Riemann surface (without boundary) $X$ can be embedded into $\mathbb P^3$.
Indeed, if $X$ has genus $g$ and $x\in X$ is an arbitrary point, the divisor $D=(2g+1)\cdot x$ is very ample and embeds $X$ into $\mathbb P^{g+1}$.
If $g\geq 2$ it is easy to project the obtained embedded Riemann surface into a linear subspace $P^3\subset \mathbb P^{g+1}$ of dimension $3$ and thus obtain the composed embedding $X\hookrightarrow P^3\cong \mathbb P^3$.

b) However Riemann surface of genus $g\geq 2$ cannot in general be embedded in the plane $ \mathbb P^2$:
one obstruction is that the genus of a smooth plane curve of degree $d$ is $g=\frac {(d-1)(d-2)}{2}$ and , of course, most integers are not of this form!
For example Riemann surfaces of genus $2,4,5,7,8,9,...$ can never be embedded into $ \mathbb P^2$ .